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kramer
3 years ago
7

Which avenue may utilize video streaming, audio narration, print designs and animation?

Computers and Technology
1 answer:
Shtirlitz [24]3 years ago
8 0

Answer:

Multimedia avenue.

Explanation:

A file type is the standard used to store data such as pictures, texts, videos, and audios. All file types have unique file extension that determine which program to use to open a particular file and to access its data e.g pictures (jpeg, png), texts (txt, docx, rtf), videos (mp4, 3gp, avi), audios (mp3, acc).

Sometimes, computer users make the mistake of opening files with the wrong software application or program, this often leads to an error due to the incompatibility of the software application with the particular file.

Basically, all software applications are designed and developed for use with specific file extensions or formats and as such, when used to open a file it isn't developed for, it result in an error.

A multimedia avenue refers to a channel that is designed and developed to accept, utilize and combine various file formats such as audio, video, text, animation effects, etc. A common example of a multimedia avenue is Microsoft PowerPoint software.

Hence, the multimedia avenue may utilize video streaming, audio narration, print designs and animation.

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g Write a program that prompts the user for an integer n between 1 and 100. If the number is outside the range, it prints an err
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Answer:

The cpp program is given below.

#include<iostream>

#include<iomanip>

using namespace std;

int main() {

   

   // variables declared

   int n;

   int sum=0;

   float avg;

   

   do

   {

       // user input taken for number    

       cout<< "Enter a number between 1 and 100 (inclusive): ";

       cin>>n;

       

       if(n<1 || n>100)

           cout<<" Number is out of range. Enter valid number."<<endl;

       

   }while(n<1 || n>100);

   

   cout<<" "<<endl;

   

   // printing even numbers between num and 50  

   for(int num=1; num<=n; num++)

   {

       sum = sum + num;

   }

   

   avg = sum/n;

   

   // displaying sum and average

   cout<<"Sum of numbers between 1 and "<<n<<" is "<<sum<<endl;

   cout<<"Average of numbers between 1 and "<<n<<" is ";

   printf("%.2f", avg);

   

       return 0;

}

OUTPUT

Enter a number between 1 and 100 (inclusive): 123

Number is out of range. Enter valid number.

Enter a number between 1 and 100 (inclusive): 56

 

Sum of numbers between 1 and 56 is 1596

Average of numbers between 1 and 56 is 28.00

Explanation:

The program is explained below.

1. Two integer variables are declared to hold the number, n, and to hold the sum of numbers from 1 to n, sum. The variable sum is initialized to 0.

2. One float variable, avg, is declared to hold average of numbers from 1 to n.

3. User input is taken for n inside do-while loop. The loop executes till user enters value between 1 and 100. Otherwise, error message is printed.

4. The for loop executes over variable num, which runs from 1 to user-entered value of n.

5. Inside for loop, all the values of num are added to sum.

sum = sum + num;

6. Outside for loop, average is computed and stored in avg.

avg = sum/n;

7. The average is printed with two numbers after decimal using the following code.

printf("%.2f", avg);

8. The program ends with return statement.

9. All the code is written inside main() and no classes are involved.

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